2.5
Applications of Linear Equations
OBJECTIVES
2.5
1. Solve linear equations when signs of grouping
are present
2. Solve applications involving numbers
3. Solve geometry problems
4. Solve mixture problems
5. Solve motion problems
In Section 1.2, we looked at the distributive property. That property is used when solving
equations in which parentheses are involved.
Let’s start by reviewing an example similar to those we considered earlier. We will then
solve other equations involving grouping symbols.
Example 1
Solving Equations with Parentheses
Solve for x:
5(2x 1) 25
First, multiply on the left to remove the parentheses, then solve as before.
NOTE Again, returning to the
original equation will catch any
possible errors in the removal
of the parentheses.
Left side
Right side
5(2 3 1) 25
5(6 1) 25
5 5 25
25 25 (True)
10x 5 25
5 5
10x
Add 5.
30
10x
30
10
10
Divide by 10.
x3
The answer is 3. To check, return to the original equation. Substitute 3 for x. Then evaluate
the left and right sides separately.
CHECK YOURSELF 1
Solve for x.
8(3x 5) 16
NOTE Given an expression such
as a (b c) you could rewrite
it as
a ((b c)).
Be especially careful if a negative sign precedes a grouping symbol. The sign of each
term inside the grouping symbol must be changed.
Example 2
© 2001 McGraw-Hill Companies
Solving Equations with Parentheses
NOTE Remember,
(3x 1) 3x 1
Change both signs.
Solve 8 (3x 1) 8.
First, remove the parentheses. The original equation then becomes
8 3x 1 8
3x 7 8
7
7
Combine like terms.
Add 7 to each side.
3x 15
x5
Divide by 3.
193
194
CHAPTER 2
EQUATIONS AND INEQUALITIES
The solution is 5. You should verify this result.
CHECK YOURSELF 2
Solve for x.
7 (4x 3) 22
Example 3 illustrates the solution process when more than one grouping symbol is
involved in an equation.
Example 3
Solving Equations with Parentheses
Solve 2(3x 1) 3(x 5) 4.
2(3x 1) 3(x 5) 4
6x 2 3x 15 4
3x 17 4
3x 21
Use the distributive property
to remove the parentheses.
Combine like terms on the left.
Add 17.
Divide by 3.
x7
The solution is 7.
To check, return to the original equation to replace x with 7.
the order of operations are
applied.
2(3 7 1) 3(7 5) 4
2(21 1) 3(7 5) 4
2 20 3 12 4
40 36 4
44
(True)
The solution is verified.
CHECK YOURSELF 3
Solve for x.
5(2x 4) 7 3(1 2x)
Many applications lead to equations involving parentheses. That means the methods of
Examples 2 and 3 will have to be applied during the solution process. Before we look at
examples, you should review the five-step process for solving word problems found in
Section 2.4.
These steps are illustrated in Example 4.
Example 4
Solving Applications Using Parentheses
One number is 5 more than a second number. If 3 times the smaller number plus 4 times the
larger is 104, find the two numbers.
© 2001 McGraw-Hill Companies
NOTE Notice how the rules for
APPLICATIONS OF LINEAR EQUATIONS
SECTION 2.5
195
Step 1 What are you asked to find? You must find the two numbers.
Step 2 Represent the unknowns. Let x be the smaller number. Then
NOTE “5 more than” x
x5
is the larger number.
Step 3 Write an equation.
3x 4(x 5) 104
NOTE Note that the
parentheses are essential in
writing the correct equation.
3 times
Plus
the smaller
4 times
the larger
Step 4 Solve the equation.
3x 4(x 5) 104
3x 4x 20 104
7x 20 104
7x 84
x 12
The smaller number (x) is 12, and the larger number (x 5) is 17.
Step 5 Check the solution: 12 is the smaller number, and 17 is the larger number.
3 12 4 17 104
(True)
CHECK YOURSELF 4
One number is 4 more than another. If 6 times the smaller minus 4 times the larger
is 4, what are the two numbers?
The solutions for many problems from geometry will also yield equations involving
parentheses. Consider Example 5.
Example 5
Solving a Geometry Application
© 2001 McGraw-Hill Companies
NOTE Whenever you are
working on an application
involving geometric figures,
you should draw a sketch of the
problem, including the labels
assigned in step 2.
Length 3 x 1
Width
x
The length of a rectangle is 1 centimeter (cm) less than 3 times the width. If the perimeter
is 54 cm, find the dimensions of the rectangle.
Step 1 You want to find the dimensions (the width and length).
Step 2 Let x be the width.
Then 3x 1 is the length.
3 times
the width
1 less than
Step 3 To write an equation, we’ll use this formula for the perimeter of a rectangle:
P 2W 2L
196
CHAPTER 2
EQUATIONS AND INEQUALITIES
So
2x 2(3x 1) 54
Twice the
width
Twice the
length
Perimeter
Step 4 Solve the equation.
2x 2(3x 1) 54
2x 6x 2 54
8x 56
NOTE Be sure to return to the
original statement of the
problem when checking your
result.
x7
The width x is 7 cm, and the length, 3x 1, is 20 cm. We leave step 5, the check, to you.
CHECK YOURSELF 5
The length of a rectangle is 5 inches (in.) more than twice the width. If the perimeter of the rectangle is 76 in., what are the dimensions of the rectangle?
You will also often use parentheses in solving mixture problems. Mixture problems involve combining things that have a different value, rate, or strength. Look at Example 6.
Example 6
Solving a Mixture Problem
Four hundred tickets were sold for a school play. General admission tickets were $4, and
student tickets were $3. If the total ticket sales were $1350, how many of each type of ticket
were sold?
Step 1 You want to find the number of each type of ticket sold.
Step 2 Let x be the number of general admission tickets.
Then 400 x student tickets were sold.
400 tickets were
sold in all.
General admission tickets:
4x
$4 for each of the x tickets
Student tickets:
3(400 x)
$3 for each of the 400 x tickets
So to form an equation, we have
4x 3(400 x) 1350
Value of
general
admission
tickets
Value of
student
tickets
Total
value
© 2001 McGraw-Hill Companies
Step 3 The sales value for each kind of ticket is found by multiplying the price of the
ticket by the number sold.
number of general admission
tickets, from 400, the total
number of tickets, to find the
number of student tickets.
NOTE We subtract x, the
APPLICATIONS OF LINEAR EQUATIONS
SECTION 2.5
197
Step 4 Solve the equation.
4x 3(400 x) 1350
4x 1200 3x 1350
x 1200 1350
x 150
So 150 general admission and 250 student tickets were sold. We leave the check to you.
CHECK YOURSELF 6
Beth bought 35¢ stamps and 15¢ stamps at the post office. If she purchased 60
stamps at a cost of $17, how many of each kind did she buy?
The next group of applications we will look at in this section involves motion problems.
They involve a distance traveled, a rate or speed, and time. To solve motion problems, we
need a relationship among these three quantities.
Suppose you travel at a rate of 50 miles per hour (mi/h) on a highway for 6 hours (h).
How far (what distance) will you have gone? To find the distance, you multiply:
NOTE Be careful to make your
(50 mi/h)(6 h) 300 mi
units consistent. If a rate is
given in miles per hour, then
the time must be given in hours
and the distance in miles.
Speed
or rate
Time
Distance
Definitions: Relationship for Motion Problems
In general, if r is a rate, t is the time, and d is the distance traveled,
drt
This is the key relationship, and it will be used in all motion problems. Let’s see how it
is applied in Example 7.
Example 7
Solving a Motion Problem
© 2001 McGraw-Hill Companies
On Friday morning Ricardo drove from his house to the beach in 4 h. In coming back on
Sunday afternoon, heavy traffic slowed his speed by 10 mi/h, and the trip took 5 h. What
was his average speed (rate) in each direction?
Step 1 We want the speed or rate in each direction.
Step 2 Let x be Ricardo’s speed to the beach. Then x 10 is his return speed.
It is always a good idea to sketch the given information in a motion problem. Here we
would have
x mi/h for 4 h
Going
x 10 mi/h for 5 h
Returning
EQUATIONS AND INEQUALITIES
NOTE Distance (going)
distance (returning)
or
Time rate (going) time rate
(returning)
Step 3 Because we know that the distance is the same each way, we can write an equation, using the fact that the product of the rate and the time each way must be the same.
So
4x 5(x 10)
Time rate
(going)
CHAPTER 2
198
Time rate
(returning)
A chart can help summarize the given information. We begin by filling in the information
given in the problem.
Distance
Going
Returning
Rate
Time
x
x 10
4
5
Now we fill in the missing information. Here we use the fact that d rt to complete the
chart.
Going
Returning
Distance
Rate
Time
4x
5(x 10)
x
x 10
4
5
From here we set the two distances equal to each other and solve as before.
Step 4 Solve.
4x 5(x 10)
4x 5x 50
x 50
x 50 mi/h
So Ricardo’s rate going to the beach was 50 mi/h, and his rate returning was 40 mi/h.
Step 5 To check, you should verify that the product of the time and the rate is the same
in each direction.
CHECK YOURSELF 7
A plane made a flight (with the wind) between two towns in 2 h. Returning against
the wind, the plane’s speed was 60 mi/h slower, and the flight took 3 h. What was
the plane’s speed in each direction?
© 2001 McGraw-Hill Companies
NOTE x was his rate going,
x 10 his rate returning.
APPLICATIONS OF LINEAR EQUATIONS
SECTION 2.5
199
Example 8 illustrates another way of using the distance relationship.
Example 8
Solving a Motion Problem
Katy leaves Las Vegas for Los Angeles at 10 A.M., driving at 50 mi/h. At 11 A.M. Jensen
leaves Los Angeles for Las Vegas, driving at 55 mi/h along the same route. If the cities are
260 mi apart, at what time will they meet?
Step 1 Let’s find the time that Katy travels until they meet.
Step 2 Let x be Katy’s time.
Then x 1 is Jensen’s time.
Jensen left 1 h later!
Again, you should draw a sketch of the given information.
(Jensen)
55 mi /h for x 1 h
(Katy)
50 mi /h for x h
Los Angeles
Las Vegas
Meeting
point
Step 3 To write an equation, we will again need the relationship d rt. From this
equation, we can write
Katy’s distance 50x
Jensen’s distance 55(x 1)
As before, we can use a table to solve.
Katy
Jensen
Distance
Rate
Time
50x
55(x 1)
50
55
x
x1
From the original problem, the sum of those distances is 260 mi, so
50x 55(x 1) 260
© 2001 McGraw-Hill Companies
Step 4
50x 55(x 1) 260
50x 55x 55 260
105x 55 260
105x 315
x3h
NOTE Be sure to answer the
question asked in the problem.
Finally, because Katy left at 10 A.M., the two will meet at 1 P.M. We leave the check of this
result to you.
CHAPTER 2
EQUATIONS AND INEQUALITIES
CHECK YOURSELF 8
At noon a jogger leaves one point, running at 8 mi/h. One hour later a bicyclist
leaves the same point, traveling at 20 mi/h in the opposite direction. At what time
will they be 36 mi apart?
CHECK YOURSELF ANSWERS
1. 1
2. 3
3. 4
4. The numbers are 10 and 14.
5. The width is
11 in.; the length is 27 in.
6. 40 at 35¢, and 20 at 15¢
7. 180 mi/h with the
wind and 120 mi/h against the wind
8. At 2 P.M.
© 2001 McGraw-Hill Companies
200
Name
2.5
Exercises
Section
Date
Solve each of the following equations for x, and check your results.
ANSWERS
1. 3(x 5) 6
2. 2(x 3) 6
1.
2.
3. 5(2x 3) 35
4. 4(3x 5) 88
3.
4.
5. 7(5x 8) 84
6. 6(3x 2) 60
5.
6.
7. 10 (x 2) 15
8. 12 (x 3) 3
7.
8.
9. 5 (2x 1) 12
10. 9 (3x 2) 2
9.
10.
11. 7 (3x 5) 13
12. 5 (4x 3) 4
11.
12.
13. 5x 3(x 6)
14. 5x 2(x 12)
13.
14.
15. 7(2x 3) 20x
16. 4(3x 5) 18x
15.
16.
17. 6(6 x) 3x
18. 5(8 x) 3x
17.
18.
© 2001 McGraw-Hill Companies
19. 2(2x 1) 3(x 1)
20. 3(3x 1) 4(2x 1)
19.
20.
21. 5(4x 2) 6(3x 4)
22. 4(6x 1) 7(3x 2)
21.
22.
23. 9(8x 1) 5(4x 6)
24. 7(3x 11) 9(3 6x)
23.
24.
201
ANSWERS
25.
25. 4(2x 1) 3(3x 1) 9
26. 7(3x 4) 8(2x 5) 13
27. 5(2x 1) 3(x 4) 4(x 4)
28. 2(x 3) 3(x 5) 3(x 2) 7
28.
29. 3(3 4x) 30 5x 2(6x 7)
30. 3x 5(3x 7) 2(x 9) 45
29.
31. 2x [3x (2x 5)] (15 2x)
30.
32. 3x [5x (x 4)] 2(x 3)
26.
27.
31.
33. 3x2 2(x2 2) x2 4
32.
34. 5x2 [2(2x2 3)] 3 x2 9
33.
Solve the following word problems. Be sure to show the equation you use for the
solution.
34.
35.
35. Number problem. One number is 8 more than another. If the sum of the smaller
number and twice the larger number is 46, find the two numbers.
36.
37.
36. Number problem. One number is 3 less than another. If 4 times the smaller number
minus 3 times the larger number is 4, find the two numbers.
38.
37. Number problem. One number is 7 less than another. If 4 times the smaller number
39.
plus 2 times the larger number is 62, find the two numbers.
40.
38. Number problem. One number is 10 more than another. If the sum of twice the
41.
smaller number and 3 times the larger number is 55, find the two numbers.
42.
39. Consecutive integers. Find two consecutive integers such that the sum of twice the
first integer and 3 times the second integer is 28. (Hint: If x represents the first
integer, x 1 represents the next consecutive integer.)
43.
40. Consecutive integers. Find two consecutive odd integers such that 3 times the first
integer is 5 more than twice the second. (Hint: If x represents the first integer, x 2
represents the next consecutive odd integer.)
its width. If the perimeter of the rectangle is 74 in., find the dimensions of the
rectangle.
42. Dimensions of a rectangle. The length of a rectangle is 5 centimeters (cm) less than
3 times its width. If the perimeter of the rectangle is 46 cm, find the dimensions of
the rectangle.
43. Garden size. The length of a rectangular garden is 4 meters (m) more than 3 times
its width. The perimeter of the garden is 56 m. What are the dimensions of the
garden?
202
© 2001 McGraw-Hill Companies
41. Dimensions of a rectangle. The length of a rectangle is 1 inch (in.) more than twice
ANSWERS
44. Size of a playing field. The length of a rectangular playing field is 5 feet (ft) less
than twice its width. If the perimeter of the playing field is 230 ft, find the length and
width of the field.
44.
45.
46.
47.
48.
49.
50.
45. Isosceles triangle. The base of an isosceles triangle is 3 cm less than the length of
the equal sides. If the perimeter of the triangle is 36 cm, find the length of each of the
sides.
46. Isosceles triangle. The length of one of the equal legs of an isosceles triangle is 3 in.
less than twice the length of the base. If the perimeter is 29 in., find the length of each
of the sides.
47. Ticket sales. Tickets for a play cost $8 for the main floor and $6 in the balcony. If
the total receipts from 500 tickets were $3600, how many of each type of ticket were
sold?
48. Ticket sales. Tickets for a basketball tournament were $6 for students and $9 for
nonstudents. Total sales were $10,500, and 250 more student tickets were sold than
nonstudent tickets. How many of each type of ticket were sold?
49. Number of stamps. Maria bought 80 stamps at the post office in 33¢ and 25¢
© 2001 McGraw-Hill Companies
denominations. If she paid $24 for the stamps, how many of each denomination did
she buy?
50. Money denominations. A bank teller had a total of 125 $10 bills and $20 bills to
start the day. If the value of the bills was $1650, how many of each denomination did
he have?
203
ANSWERS
51.
51. Ticket sales. Tickets for a train excursion were $120 for a sleeping room, $80 for a
52.
berth, and $50 for a coach seat. The total ticket sales were $8600. If there were 20
more berth tickets sold than sleeping room tickets and 3 times as many coach tickets
as sleeping room tickets, how many of each type of ticket were sold?
53.
52. Baseball tickets. Admission for a college baseball game is $6 for box seats, $5 for
the grandstand, and $3 for the bleachers. The total receipts for one evening were
$9000. There were 100 more grandstand tickets sold than box seat tickets. Twice as
many bleacher tickets were sold as box seat tickets. How many tickets of each type
were sold?
54.
55.
56.
53. Driving speed. Patrick drove 3 hours (h) to attend a meeting. On the return trip, his
speed was 10 miles per hour (mi/h) less and the trip took 4 h. What was his speed
each way?
57.
58.
54. Bicycle speed. A bicyclist rode into the country for 5 h. In returning, her speed was
59.
5 mi/h faster and the trip took 4 h. What was her speed each way?
55. Driving speed. A car leaves a city and goes north at a rate of 50 mi/h at 2 P.M. One
hour later a second car leaves, traveling south at a rate of 40 mi/h. At what time will
the two cars be 320 mi apart?
56. Bus distance. A bus leaves a station at 1 P.M., traveling west at an average rate of
44 mi/h. One hour later a second bus leaves the same station, traveling east at a rate
of 48 mi/h. At what time will the two buses be 274 mi apart?
57. Traveling time. At 8:00 A.M., Catherine leaves on a trip at 45 mi/h. One hour later,
58. Bicycling time. Martina leaves home at 9 A.M., bicycling at a rate of 24 mi/h. Two
hours later, John leaves, driving at the rate of 48 mi/h. At what time will John catch
up with Martina?
59. Traveling time. Mika leaves Boston for Baltimore at 10:00 A.M., traveling at 45 mi/h.
One hour later, Hiroko leaves Baltimore for Boston on the same route, traveling at
50 mi/h. If the two cities are 425 mi apart, when will Mika and Hiroko meet?
204
© 2001 McGraw-Hill Companies
Max decides to join her and leaves along the same route, traveling at 54 mi/h. When
will Max catch up with Catherine?
ANSWERS
60. Traveling time. A train leaves town A for town B, traveling at 35 mi/h. At the same
time, a second train leaves town B for town A at 45 mi/h. If the two towns are 320 mi
apart, how long will it take for the two trains to meet?
61. Tree inventory. There are 500 Douglas fir and hemlock trees in a section of forest
bought by Hoodoo Logging Co. The company paid an average of $250 for each
Douglas fir and $300 for each hemlock. If the company paid $132,000 for the trees,
how many of each kind did the company buy?
60.
61.
62.
63.
62. Tree inventory. There are 850 Douglas fir and ponderosa pine trees in a section of
forest bought by Sawz Logging Co. The company paid an average of $300 for each
Douglas fir and $225 for each ponderosa pine. If the company paid $217,500 for the
trees, how many of each kind did the company buy?
64.
65.
66.
63. There is a universally agreed on “order of operations” used to simplify expressions.
Explain how the order of operations is used in solving equations. Be sure to use
complete sentences.
64. A common mistake when solving equations is the following:
The equation:
First step in solving:
2(x 2) x 3
2x 2 x 3
Write a clear explanation of what error has been made. What could be done to avoid
this error?
65. Another very common mistake is in the equation below:
The equation:
First step in solving:
6x (x 3) 5 2x
6x x 3 5 2x
© 2001 McGraw-Hill Companies
Write a clear explanation of what error has been made and what could be done to
avoid the mistake.
66. Write an algebraic equation for the English statement “Subtract 5 from the sum of x
and 7 times 3 and the result is 20.” Compare your equation with other students. Did
you all write the same equation? Are all the equations correct even though they don’t
look alike? Do all the equations have the same solution? What is wrong? The English
statement is ambiguous. Write another English statement that leads correctly to more
than one algebraic equation. Exchange with another student and see if they think the
statement is ambiguous. Notice that the algebra is not ambiguous!
205
ANSWERS
a.
Getting Ready for Section 2.6 [Section 1.4]
b.
Evaluate the following.
c.
d.
(a)
270 10
90
(b)
660 100
11
(c)
120 100
4000
e.
(d)
320 100
2.5
(e)
23 4.5
10
(f )
46 15
100
f.
Answers
1. 7
3. 2
15. 29.
39.
47.
51.
57.
63.
7
2
17. 4
5. 4
19. 5
7. 3
21. 7
11. 9. 4
23.
3
4
1
3
25. 2
13. 9
27. 3
5
31. 2
33. All real numbers
35. 10, 18
37. 8, 15
5, 6
41. 12 in., 25 in.
43. 6 m, 22 m
45. Legs, 13 cm; base, 10 cm
200 $6 tickets, 300 $8 tickets
49. 30 25¢ stamps, 50 33¢ stamps
60 coach, 40 berth, and 20 sleeping room
53. 40 mi/h, 30 mi/h
55. 6 P.M.
2 P.M.
59. 3 P.M.
61. 360 Douglas fir, 140 hemlock
65.
a. 30
b. 6000
c. 3
d. 12,800
f. 6.9
© 2001 McGraw-Hill Companies
e. 10.35
206